### Vector and Matrix

Exercise Express the vectors **b** =
(1, 1, 1), **c** = (-1, 0, 1), **e**_{1} = (1, 0,
0) as linear combinations of the vectors **a**_{1} = (1, 2,
3), **a**_{2} = (4, 5, 6), **a**_{3} = (7,
8, 9). Is the expression (i.e., the coefficients) unique?

Answer We would like to write **b**,
**c** , **e**_{1} in the form `x`_{1}(1,
2, 3) + `x`_{2}(4, 5, 6) + `x`_{3}(7, 8, 9). The computation
is similar to the previous exercise.
From this exercise,
we have

`x`_{1}[ |
1 |
] + `x`_{2}[ |
4 |
] + `x`_{3}[ |
7 |
] = [ |
`x`_{1} + 4`x`_{2} + 7`x`_{3} |
] |

2 |
5 |
8 |
2`x`_{1} + 5`x`_{2} + 8`x`_{3} |

3 |
6 |
9 |
3`x`_{1} + 6`x`_{2} + 9`x`_{3} |

Therefore the problem is to solve the system.

`x`_{1} |
+ 4`x`_{2} |
+ 7`x`_{3} |
= |
`b`_{1} |

2`x`_{1} |
+ 5`x`_{2} |
+ 8`x`_{3} |
= |
`b`_{2} |

3`x`_{1} |
+ 6`x`_{2} |
+ 9`x`_{3} |
= |
`b`_{3} |

For `b`_{1} = `b`_{2} = `b`_{3} = 1
on the right side, we find `x`_{1} = -1/3, `x`_{2}
= 1/3, `x`_{3} = 0 is one solution. Therefore ** **`b`
= -1/3 **a**_{1} + 1/3 **a**_{2}. Since
the solution is not unique, the expression is not unique. Another possibility
is ** **`b` = -1/6 **a**_{1} + 1/6 **a**_{3}.

For `b`_{1} = -1, `b`_{2} = 0, `b`_{3}
= 1 on the right side, we find `x`_{1} = 4/3, `x`_{2}
= 0, `x`_{3} = -1/3 is one solution. Therefore ** **`c`
= 4/3 **a**_{1} - 1/3 **a**_{3}. Again the
expression is not unique.

For `b`_{1} = 1, `b`_{2} = 0, `b`_{3}
= 0 on the right side, there is no solution. Therefore ** ****e**_{1}
is not a linear combination of **a**_{1, }**a**_{2,
} **a**_{3}.